Question

The speed of a 2 kg mass on a spring is 5 m/s as it passes through its equilibrium position. What is its frequency if the amplitude is 2.5 m

Answer 1

Given :

Mass of the object attached to the spring, m = 2 kg

Velocity of the object as it moves, V = 5 m/s

Amplitude of the object as it swings, A = 2.5 m

We have to find the frequency of the object.

We known,

`$V_(max) = \omega A = 2 \pi f A$`

Therefore,
`$f= (V_(max))/(2 \pi A)$`

`$f= (5)/(2 * 3.14 * 2.5)$`

f = 0.32 Hz

Therefore the frequency of the object is 0.32 hertz when the amplitude of the 2 kg mass is 2.5 m moving a speed of 5 m/s.